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Intro to Analysis Proof help

Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.

I've attached the problem.

for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer

for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2

For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!

Re: Intro to Analysis Proof help

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Originally Posted by

**glambeth** Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.

I've attached the problem.

for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer

Then you have misunderstood the whole thing! The theorem you are trying to prove is that "If p is the not the square of an integer and $\displaystyle x^2= p$ then x is irrational". You cannot just **choose** a value for x. x is "given" by $\displaystyle x^2= p$. The crucial point here is "Assume that x is a positive rational number. What is the **definition** of "rational number"?

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for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2

Again, you cannot **assume** a value for x. x is given by $\displaystyle x^2= p$ and you must use that x.

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For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!

Re: Intro to Analysis Proof help

Quote:

Originally Posted by

**glambeth** Hi, I'm self studying analysis, and I'm having problem with this proof. I've gotten most in my book so far but this has posed a problem.

I've attached the problem.

for part 1 I just said A is non empty since when x = 1/5 and n = 5 1 is an integer

for part 2 I let k = 1, x = 3/2 and showed that 1 < 3/2 < 2

For part 3 and 4 I have no idea what to do and i'm not sure whether i've done the prior correctly. I'd appreciate any help. Thanks!

Do not use examples in proofs.

Because $\displaystyle x\in\mathbb{Q}$ then there is an integer $\displaystyle n$ such that $\displaystyle nx\in J$ so $\displaystyle A\ne\emptyset.$

Use well-ordering to say that $\displaystyle A$ has a first term call it $\displaystyle m$.

Again $\displaystyle x$ cannot be integer so it is between two consecutive integers, $\displaystyle k<x<k+1$ or

$\displaystyle \\0<x-k<1\\0<mx-mk<m$.

You know that both $\displaystyle mx~\&~mk$ are integers (WHY?) thus $\displaystyle (mx-mk)x$ is also an integer.

But $\displaystyle (mx-mk)\notin A$ so there is your contradiction . (HOW & WHY??)

Re: Intro to Analysis Proof help

Wow! I misread that problem completely, I appreciate the help. I'll let you guys know if I have any problems with constructing my proof! Thanks again!

Re: Intro to Analysis Proof help

Could you explain how mx and mk are integers and why the last line is a contradiction?